How do you solve Count Prefix and Suffix Pairs I on LeetCode?

Count Prefix and Suffix Pairs I (LeetCode #3042) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n) space complexity. Key insight: Understanding prefix and suffix relationships is crucial.

What is the brute force solution for Count Prefix and Suffix Pairs I?

The brute force approach for Count Prefix and Suffix Pairs I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of strings in the array to see if one is both a prefix and a suffix of the other. This is straightforward but can be slow for larger inputs. The optimal Optimal Solution approach improves this to O(n * m).

What are the interview tips for Count Prefix and Suffix Pairs I?

Explain your thought process clearly as you code. Consider edge cases and test your solution with them. Discuss time and space complexity before finalizing your approach.

What are common mistakes when solving Count Prefix and Suffix Pairs I?

Not considering the case where the prefix and suffix are the same word. Overlooking edge cases with empty strings or single-character strings.

#3042

Count Prefix and Suffix Pairs I

Easy

ArrayStringTrieRolling HashString MatchingHash FunctionHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n)

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Intuition

Time O(n * m)Space O(n)

The optimal solution uses a HashMap to store the words and their lengths, allowing us to quickly check if a word can be a prefix or suffix of another. This reduces the number of comparisons needed.

⚙️

Algorithm

4 steps

1Step 1: Create a HashMap to store the words and their lengths.
2Step 2: Iterate through each word and check for all possible lengths of prefixes that could also be suffixes in other words.
3Step 3: For each word, check if it exists in the HashMap with the required conditions.
4Step 4: Count valid pairs and return the total.

solution.py11 lines

1def countPairs(words):
2    word_map = {}
3    count = 0
4    for word in words:
5        word_map[word] = len(word)
6    for word in words:
7        for length in range(1, len(word) + 1):
8            prefix = word[:length]
9            if prefix in word_map and prefix != word:
10                count += 1
11    return count

ℹ

Complexity note: The time complexity is O(n * m) where n is the number of words and m is the average length of the words. The space complexity is O(n) due to the HashMap storing the words.

1Understanding prefix and suffix relationships is crucial.
2Utilizing data structures like HashMap can optimize search operations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.